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Edvantage Science AP®CHEMISTRY 2

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m=0 y = 100 k = 27

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AP Chemistry 2 Copyright © 2015, Edvantage Interactive

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Authors Cheri Smith Yale Secondary School District 34 Abbotsford Gary Davidson School District 22 Vernon Megan Ryan Walnut Grove Secondary School District 35 Langley Chris Toth St. Thomas More Collegiate Burnaby Program Consultant Lionel Sandner Edvantage Interactive

The AP Big Ideas at the beginning of each chapter are quoted from AP Chemistry: Course and Exam Description, revised edition, effective Fall 2013, published by the College Board, New York, NY. Advanced Placement, AP, and College Board are registered trademarks of the College Board.

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3 Thermodynamics This chapter focuses on the following AP Big Idea from the College Board:

•

Big Idea 5: The laws of thermodynamics describe the essential role of energy and explain and predict the direction of changes in matter.

By the end of this chapter, you should be able to do the following:

• • • •

Use representations and models to predict the sign and relative magnitude of the entropy change associated with chemical or physical processes Predict, either quantitatively or qualitatively, whether or not a physical or chemical process is thermodynamically favored by determination of the signs of both ΔH° and ΔS°, and calculation or estimation of ΔG° when needed Determine whether a chemical or physical process is thermodynamically favorable by calculating the change in standard Gibbs free energy Explain why a thermodynamically favored chemical reaction may not produce large amounts of product based on consideration of both initial conditions and kinetic effects, or why a thermodynamically unfavored chemical reaction can produce large amounts of product for certain sets of initial conditions

By the end of this chapter, you should know the meaning of these key terms:

• • • • • • • • •

endergonic exergonic Gibbs free energy microstates positional disorder second law of thermodynamics thermal disorder thermodynamically favorable third law of thermodynamics

Thermodynamics is the study of heat and temperature and their relation to energy and work.

It includes concepts such as entropy and pressure.

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3.1 Entropy — A Quantitative Treatment Warm Up Section 2.4 (Entropy Change versus Enthalpy Change) dealt with the impact of ΔS and ΔH on spontaneity. In AP Chemistry 1, we learned that heat flows from a region of high energy to a region of low energy. This might lead to the conclusion that spontaneous change must be exothermic.

1. Consider the following change occurring at room temperature:

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(a) Is this change spontaneous (did it occur without outside intervention)? _______ (b) Is this process endothermic or exothermic? ________________ (c) What is the sign of ΔH? _______ 2.

Consider the change at a molecular level: �����

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(a) What is the sign for ΔS? _______ (b) Is this process enthalpy driven or entropy driven? ______________ Explain. ____________________________________________________________________________________________ ___________________________________________________________________________________________________

The First Law of Thermodynamics

Spontaneous Change

Energy is conserved — never created or destroyed. When energy is released by an exothermic change, the potential energy stored in chemical bonds is converted to kinetic energy, which is the energy of motion of atoms and molecules. As the kinetic energy of the atoms and molecules in the system and the surroundings increases, the temperature rises. Sometimes we speak of exothermic changes as “giving off heat.” Heat is the energy transferred between a system and its surroundings.

When thermodynamicists say that a reaction is spontaneous, they mean that the reaction will occur without any outside intervention. They do not mean the reaction occurs rapidly. We studied the rate or speed of a reaction in Unit 1: Reaction Kinetics.

In contrast, a non-spontaneous event can continue only as long as it receives some sort of outside assistance. All non-spontaneous events occur at the expense of spontaneous ones. They can occur only when some spontaneous change occurs first.

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Enthalpy and Spontaneity

Figure 3.1.1 Every

spontaneous process increases the entropy of the universe.

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When a change lowers the potential energy of a system, it tends to occur spontaneously. Fuels evolve heat as they combust to form H2O and CO2 and convert the potential energy in the bonds of the reactants to heat. Since a change that lowers the potential energy of a system is exothermic, we can say that exothermic changes have a tendency to occur spontaneously.

While many spontaneous changes are indeed exothermic, there are common examples of spontaneous endothermic changes including the melting of ice, the vaporization of water, and the dissolving of most salts in water to form aqueous solutions. Since each of these events is both endothermic AND spontaneous, they must have some other factor in common that allows them to overcome the unfavorable energy change. Careful examination shows that in each of these events, there is an increase in the randomness or disorder of the system. These observations led to the development of the second law of thermodynamics.

Second law of thermodynamics: The entropy of the universe increases with every spontaneous process.

Quick Check 1.

Use qualitative reasoning to decide which of the two options in each of these examples will occur spontaneously: (a) A mole of ice at 0°C or a mole of liquid water at the same temperature

(b) Solid sodium chloride in the bottom of a bucket of water or a solution of the salt in water

(c) A collection of jigsaw pieces or the completed puzzle

(d) One kilogram of raw rubber or 1 kg of vulcanized rubber (Vulcanized rubber is rubber that is polymerized into long chains; vulcanized rubber is used to make tires.)

(e) A pack of cards arranged in suits or a pack of cards that is randomly shuffled

The Boltzmann Equation

Ludwig Boltzmann, a famous physicist, studied fluid mechanics in the late 1800s. During this time, Boltzmann determined the probability that a group of particles might occupy a particular set of positions in space at any particular time. These sets of positions are referred to as microstates. It has been suggested that this term, microstates (states), may have led to the symbol S being selected to represent entropy. Boltzmann noted that entropy is directly proportional to the natural log of the number of microstates available. In 1872, Boltzmann derived his famous equation to represent entropy: S = kB ln W where S = entropy kB = Boltzmann’s constant = R/NA = 8.314 L·kPa/mol·K = 1.38 × 10–23 J/K 6.02 × 1023 mol–1 W = number of microstates

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In a pure crystal at absolute zero (0 K), the particles orient themselves in only one possible microstate. At absolute zero, the entropy is S = 1.38 x 10–23 J/K (ln 1). Since the natural log of 1 = 0, the entropy of a pure crystal at 0 K equals zero. This is the third law of thermodynamics.

Third law of thermodynamics: The entropy of a pure crystal at absolute zero is zero (S = 1.38 x 10–23 J/K × ln (1) = 0).

In section 2.4, we explained that increasing the number of electrons in an element or increasing the number of atoms in a molecule leads to an increase in entropy. More electrons or atoms mean more possible microstates and therefore a higher entropy according to the Boltzmann equation.

Dependence of Entropy on Temperature

The second law of thermodynamics states “the entropy of the universe increases with every spontaneous change.” The “universe” is the system plus the surroundings and the entropy of a system is a quantitative measure of the disorder or randomness within a system. There are two kinds of disorder in a system: positional disorder, depending on the arrangement of particles (molecules and atoms) in space, and thermal disorder, depending on how the total energy is distributed among all the particles. The more random the distribution of particles in space, the higher the entropy. The more randomly energy is distributed among the particles, the greater the entropy.

Table 3.1.1 shows all possible arrangements for two identical particles among three relative energy levels. The most likely total energy is 2 kJ. Maximizing positional disorder does not necessarily lead to maximum overall energy.

Table 3.1.1 All possible two-particle arrangements among three relative energy levels

Energy Level

Particle Arrangements

2 kJ

X

1 kJ

X

0 kJ

XX

X

X

Total energy

0 kJ

1 kJ

2 kJ

X XX

X

2 kJ

3 kJ

XX

4 kJ

Adding heat to a system results in greater disorder and, therefore, greater entropy. The entropy increase is directly proportional to the amount of heat added. However, the increase in entropy is more significant when we add heat to a system at a lower temperature where little disorder exists, rather than at high temperature where there is already substantial disorder. For a given quantity of heat, the entropy change is inversely proportional to the temperature at which we add the heat. Thus entropy can be determined as: Ssurroundings = q = –ΔH T T

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The equation above shows that entropy has units of energy divided by temperature (J/K). If an entropy change is being calculated for a particular quantity of material undergoing a physical or chemical change, the units may be J/K·mol.

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������� Figure 3.1.2 The change in entropy (ΔS) is most noticeable at a low temperature.

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Quick Check 1.

When 1.0 mole of steam condenses to form 1.0 mole of liquid water at 100.0° C, 40.7 kJ of heat is released. What is ΔS for the surroundings?

2.

Select the example with greater entropy in each case. (a) F2(g) or Cl2(g)

(b) CO2(g) or CO(g) (c)

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Entropy Is a State Function

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We know the third law of thermodynamics states that at absolute zero, the entropy of a pure crystal is equal to zero joules. Mathematically, we can state: S = 0 J when W = 1 at T = 0 K Because we know the point at which entropy has a value of 0 J, it is possible to determine the actual amount of entropy that a substance possesses at temperatures above 0 K. If we determine the entropy of 1 mole of a substance at a temperature of 298 K (25°C) and a pressure of 1 atm, we call it the standard entropy or S°. Tables of standard entropies are available, similar to the tables of standard enthalpies of formation used in Table 4.5.1 in AP 1. These values are listed in Table A8 in the appendix at the back of this book. (Notice that the standard entropy values of pure elements in their standard states are not equal to zero because standard entropy values are not recorded at 0 K.) Once we know the entropies of a variety of substances we can calculate the standard entropy change, or ΔS °, for chemical reactions in the same way as we calculated ΔH ° earlier.

Entropy is a state function — it is not pathway dependent. Therefore, we can use the same logic as that of a Hess’s law type of calculation to determine the entropy change for a given chemical reaction. It is important to note that like enthalpy, entropy is an extensive property — it depends on the amount of substance present. This means that the number of moles of all species in the reaction must be taken into account. Hence:

ΔS reaction ° = ∑ nS products ° – ∑ nS reactants °

Recall that for a process to be spontaneous, there must be an overall increase in the entropy of the universe. If the entropy change of the universe is negative, the process is non-spontaneous. A spontaneous entropy change is positive or in the reverse direction of non-spontaneous change.

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Sample Problem 3.1.1 — Quantifying Entropy Change Urea (from urine) hydrolyzes slowly in the presence of water to produce ammonia and carbon dioxide.

CO(NH2)2(aq) + H2O(l) → CO2(g) + 2 NH3(g) What is the standard entropy change, for this reaction when 1 mol of urea reacts with water? Is the process spontaneous? The standard entropies for reactants and products are: Substance S° (J mol–1 K–1) Substance S° (J mol–1 K–1) 173.7 CO2(g) 213.8 CO(NH2)2(aq) 70.0 NH3(g) 192.8 H2O(l)

What to Think About 1.

Calculate the change in entropy using the equation: ΔS°rxn = ∑nS°products – ∑nS°reactants

2.

After determining the stoichiometrically correct entropy values, substitute and solve.

3.

Does the answer make sense?

One mole of aqueous substance and 1 mole of liquid form 3 moles of gas.

4.

What does the answer mean?

How to Do It

Determine So values for all reactants and products. These are listed in the table given above in this example.

[mol rxn mol K mol rxn mol K] [ mol rxn mol K mol rxn mol K ] 1 mol CO2 x 213.8 J + 2 mol NH3 x 192.8 J

_

1 mol CO(NH2)2 x 173.7 J + 1 mol H2O x 70.0 J

= 599.4 – 243.7 = 355.7 J/molrxn·K The number of microstates (W) has increased.

So we expect a positive ∆S value. ✓ A positive ∆S value is indicative of a spontaneous change

Practice Problems 3.1.1 — Quantifying Entropy Change 1.

Use qualitative reasoning (no calculations are necessary) to predict whether the entropy change in each of the following processes is positive or negative. Does the entropy change of the system favor spontaneous change or not? (a) The evaporation of 1 mole of rubbing alcohol (b) 2 Ca(s) + O2(g) → 2 CaO(s) (c) XeF4(g) → Xe(g) + 2 F2(g) (d) The dilution of an aqueous solution of lime Kool-Aid

2.

Use the standard entropy table in Table A8 in the appendix at the back of this book to calculate the standard entropy change for each of the following reactions. Is the reaction thermodynamically favorable (spontaneous) based on the ΔS o value? (a) CaO(s) + 2 HCl(g) → CaCl2(s) + H2O(l)

(b) C2H4(g) + H2(g) → C2H6(g)

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3.1 Review Questions 1.

Predict the sign you would expect for ΔS° in each of the following processes. Give a short explanation of why you think it is (+) or (–). Do NOT do any calculations. (a) Blowing up a building

(b) Organizing a stack of files in alphabetical order

(c) AgCl(s) → Ag+(aq) + Cl–(aq)

(d) 2 H2(g) + O2 (g) ➝ 2 H2O(l)

(e) Na(s) + ½ Cl2(g) ➝ NaCl(s)

(f ) Removing medication from bottles and organizing the pills by the day in “blister packs”

2.

Predict the sign of ΔS° you would expect for each of the following changes. Give a short explanation of why you think it is (+) or (–). Do NOT do any calculations.

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3.

The boiling point of chloroform (CHCl3) is 61.7° C. The enthalpy of vaporization is 31.4 kJ/mol. Calculate the entropy of vaporization. Does the sign for entropy of vaporization make sense? Explain.

4.

(a)

What is the normal boiling point for formic acid, HCOOH? Given: ΔH°f = –424.72 kJ/mol for liquid HCOOH and –379.1 kJ/mol for gaseous HCOOH S ° = 128.95 J/mol K for liquid HCOOH and 251.0 J/mol K for gaseous HCOOH

(b)

Based on the calculated boiling point, in what state does formic acid exist at room temperature? Explain.

5.

The normal boiling point of water is 100.0°C , and its molar enthalpy of vaporization is 40.7 kJ/mol. What is the change in entropy in the system in J/K when 39.3 g of steam at 1 atm condenses to a liquid at the normal boiling point? Does the sign for the value you calculate make sense? Explain.

6.

° of Iodine is an unusual element that sublimes under standard conditions. If the ΔS subl ° of iodine is 145 J/mol K and the ΔH subl iodine is 62 kJ/mol, what is the standard sublimation temperature for iodine?

7.

Consider the following addition reaction: C2H2(g) + 2 H2(g) → C2H6(g) The following data is known for this reaction: Substance

S° (J/mol K)

ΔHf (kJ/mol)

C2H2(g)

200.9

226.7

H2(g)

130.7

0

C2H6(g)

????

–84.7

(a) If the value of ΔS ° for the reaction is –232.7 J/mol K, calculate the value of S ° for C2H6 gas.

(b) Calculate the enthalpy change ΔH ° for the reaction (see section 4.4 from AP I for review if needed).

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(c) Assume that Boltzmann’s equation (page 183) applies when the system is in equilibrium. What is the temperature at this point (use the non-SI unit of °C)?

8.

Consider the following reaction: 2 H2S(g) + SO2(g) → 3 S(s) + 2 H2O(g) (a) Predict the sign of ΔS° for the reaction at 298 K. Explain the basis of your prediction.

(b) Calculate the actual value of ΔS° for the reaction. Assume the sulphur produced is the rhombic form. Does your calculation justify your prediction? Explain.

9.

The combustion of methanol is described by the following equation: 2 CH3OH(l) + 3 O2(g) → 4 H2O(l) + 2 CO2(g) The value of ∆S° for this reaction is –38.6 cal/mol K at 25°C. One calorie is the energy required to warm 1 g of water by 1°C. One calorie is equivalent to 4.18 J. Data collected for this reaction: Reaction Species

∆H° (kcal/mol) at SATP

S° (cal/mol K) at SATP

CH3OH(l)

–57.0

30.3

H2O(l)

–68.3

16.7

CO2(g)

–94.0

51.1

f

Calculate the standard absolute entropy S° per mole of O2(g). Convert this value to units of J/mol K.

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10. Use the values in Table A8 in the appendix to calculate ΔS ° for each of the following equations. (It is important to use the values in the appendix because entropy values may differ slightly from one source to another. It is important to refer to the recommended source for reference values to ensure consistency in answers.) (a) H2(g) + ½ O2(g) → H2O(g)

(b) 3 O2(g) → 2 O3(g)

(c) CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g)

(d) P4O10(s) + 6 H2O(l) → 4 H3PO4(s) (Given S° for solid phosphoric acid = 110.54 J/mol K.)

(e) The complete combustion of octane in gasoline

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3.2 Gibbs Free Energy Warm Up ��

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1.

Write a balanced chemical equation (including phase subscripts) for the reaction shown in the diagram.

2.

Use the values in Table A8 in the appendix at the back of this book to calculate ΔHrxn and ΔSrxn for the reaction.

3.

What is the function of the candle in this process?

________________________________________________________________________________________________________ 4.

Is this spontaneous process enthalpy driven or entropy driven? Explain.

________________________________________________________________________________________________________ ________________________________________________________________________________________________________

Free Energy — Enthalpy and Entropy Combine

We have now seen that two factors — enthalpy and entropy — determine whether or not a physical or chemical event is spontaneous or thermodynamically favorable. Sometimes these two factors work together. For example, when a stone wall crumbles, its enthalpy decreases and its entropy increases.

Since a decrease in enthalpy and an increase in entropy both favor a spontaneous change, the two factors complement one another. In other situations, the effects of enthalpy and entropy are in opposition. Such is the case in the melting of ice, the vaporization of water and even the formation of water from its elements as in the Warm Up. When enthalpy and entropy effects oppose one another, the overall reaction spontaneity is less obvious. We can determine the net effect of the two factors through another thermodynamic quantity called Gibbs free energy, G, named after Josiah Willard Gibbs (1839–1903), a U.S. scientist (Figure 3.2.1). Gibbs free energy is defined as: G = H – TS. For a chemical or physical change at constant temperature and pressure, the equation becomes: ΔG = ΔH – TΔS

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A change is spontaneous only if it is accompanied by a decrease in free energy. In other words, ΔG must be negative for a spontaneous change. A spontaneous change is also termed a thermodynamically favorable change.

Spontaneous change must be accompanied by a decrease in free energy. That is, ΔG must be NEGATIVE for a spontaneous change.

Table 3.2.1 shows the effects of positive and negative values of ΔH and ΔS on ΔG. The table shows the importance of the absolute temperature in determining whether a change will occur spontaneously. Temperature is important in cases where ΔH and ΔS have the same signs.

Table 3.2.1 Impact of ΔH and ΔS at various temperatures on the thermodynamic favorability (spontaneity) of

chemical reactions ΔH

ΔS

−

+

Always negative

+

−

Always positive

−

−

+

+

Negative at low temps; positive at high temps Positive at low temps; negative at high temps

Figure 3.2.1 American

scientist Josiah Willard Gibbs

ΔG

Reaction Characteristics Reaction is spontaneous at all temperatures; reverse reaction is always non-spontaneous.

Reaction is non-spontaneous at all temperatures; reverse reaction occurs spontaneously.

Reaction is spontaneous at low temperatures and becomes nonspontaneous as temperature is raised.

Examples 2 O3(g) → 3O2(g)

3 O2(g) →2 O3(g)

CaO(s) + CO2(g) → CaCO3(s)

Reaction is non-spontaneous at low temperatures but becomes spontaneous as temperature is raised.

CaCO3(s) → CaO(s) + CO2(g)

Reactions that occur with a free energy decrease in the system are exergonic. Those that occur with a free energy increase are endergonic. ������������������������� �����������������������

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Figure 3.2.2 Exergonic (left) and endergonic reactions

The terms exergonic and endergonic are most commonly applied in biological contexts (Figure 3.2.2). Catabolic processes break down large proteins and lipids into amino and fatty acids producing many small molecules, each resulting in a large number of bond formations. These catabolic processes are exergonic. Anabolic processes form biological macromolecules following a multitude of bond breakages. Anabolic processes are endergonic. Essentially, anabolic processes are “powered” by catabolic processes during the metabolic processes that occur in our bodies. When one reaction is “powered” by another, the reactions are said to be coupled.

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Free Energy and Maximum Work

Free Energy and Equilibrium

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Spontaneous chemical reactions perform useful work. The burning of fuels in heat engines, chemical reactions in batteries, and the catabolic portion of metabolism within our bodies are all spontaneous processes that produce useful work. We cannot harness all of the energy released when spontaneous chemical reactions occur. Some energy is lost as heat. Engineers seek to maximize the efficiency of converting chemical energy to work and to minimize the amount of energy transferred unproductively to the environment as heat.

Is there a limit to the amount of energy in a reaction that can be harnessed as useful work? The maximum amount of energy we can harness from a reaction, as work is equal to ΔG. This is the energy that is not lost as heat and is therefore free (or available) to be used for work. By determining the value of ΔG, we can find out whether or not a given reaction will be an effective source of energy. By comparing the actual work derived from a given system with the ΔG value for the reactions involved, we can measure the efficiency of the system.

Reactions with a negative free energy change are spontaneous (thermodynamically favorable), while those with a positive free energy change are non-spontaneous. (The reverse reaction is spontaneous.)

When the value of ΔG is equal to zero, the reaction is at equilibrium. We learned in chapter 2 that at equilibrium the forward and reverse reactions occur at equal rates. A system is at equilibrium during a state change.

For the state change, H2O(l) H2O(s), equilibrium exists at 0°C and atmospheric pressure. Above 0°C, only liquid water can exist, and below 0°C, all the liquid will freeze to form ice. This gives us an interesting relationship between ΔH and ΔS, which we examined in section 3.1. ΔG = 0 = ΔH − TΔS Therefore at equilibrium, ΔH = TΔS; and ΔSsystem = ΔH T Thus, if we know ΔH and ΔS, we can calculate the temperature at which equilibrium will occur.

This is the temperature at which a state change occurs.

Quick Check 1.

For the process Br2(l)  Br2(g), ΔH = +31.0 kJ/mol and ΔS = 92.9 J/mol K. (a) Explain why a (+) value for the enthalpy change is reasonable for this process. _______________________________________________________________________________________________ _______________________________________________________________________________________________ (b) Why would we expect the entropy change to be (+) for this process? _______________________________________________________________________________________________ (c) Assuming that ΔH and ΔS are temperature independent, calculate the temperature at which Br2(l) will be in equilibrium with Br2(g). Give the answer in °C.

(d) What property does this temperature represent for bromine? _______________________________________________________________________________________________

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Standard Free Energies

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About the same time the American Josiah Gibbs derived the free energy equation, German physicist Hermann von Helmholtz deduced the same thing. For this reason, the Gibbs-Helmholtz equation is a common name for ΔG = ΔH − TΔS. The equation is valid under all temperature, pressure, and concentration conditions. We will begin by restricting all of our calculations to standard conditions of pressure (1 atm) and concentration (1 M, where it applies). In other words, we will use the equation in the form: ΔG° = ΔH° − TΔS

Where ΔG° is the standard free energy change (determined at 1 atm, 1 M), ΔH° is the standard enthalpy change calculated from reference values, ΔS° is the standard entropy change calculated from reference values, and T is the Kelvin temperature. ����������

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Figure 3.2.3 This graph of energy vs. temperature shows that the difference between enthalpy (H) and free

energy (G) is equal to TS. The graph shows that as the temperature increases, the difference between these two forms of energy becomes larger.

Since enthalpy and entropy are state functions, it follows that Gibbs free energy is also a state function. We can use Hess’s law and we can calculate standard free energies of formation as follows: ΔG° = ∑ nΔG°prod – ∑ nΔG°react

Note: ΔG°f = 0 for an element in its normal state at 25°C and 1 atm.

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Sample Problem 3.2.1(a) — Using the Gibbs-Helmholtz Equation to Calculate a Free Energy Change at Standard Temperature For the process NaCl(s) → Na+(aq) + Cl–(aq), use the tables in Table A8 in the appendix at the back of this book to calculate: (a) ΔH° (b) ΔS° (c) Use the values from (a) and (b) to calculate ΔG °. (d) According to your results, is this a spontaneous process?

What to Think About 1.

How to Do It

As ΔH° is a state function, calculate it using the equation introduced in section 4.5 of AP 1. The ΔH°f values are listed in Table A8 in the appendix.

a) ∆H°(overall) = Σn∆H°f(prod) – Σn∆H°f(react)

Notice that while elements in their standard states have ΔH°f values of 0 kJ/mol, this is not the case for ions in solution.

∆H°f for NaCl(s) = –411.15 kJ/mol

∆H°f for Na+(aq) = –240.12 kJ/mol ∆H°f for Cl–(aq) = –167.16 kJ/mol

[ [

(

)

(

)]

– 1 mol NaCl –411.15 kJ mol rxn mol NaCl

2.

We use a similar process to determine ΔS°, also a state function. The process is similar to examples shown in section 3.1 in AP 2.

Absolute entropy values of formation S°f are in Table A8 in the appendix.

)]

(

+ + 1 mol Cl- –167.16 kJ ∆H° = 1 mol Na –240.12 kJ + mol rxn mol Clmol rxn mol Na

(b) ∆S° = ΣnS°prod – ΣnS°react

= 3.87 kJ/mol rxn

S°f for Na+(aq) = 59.0 J/K mol

S°f for Cl-(aq) = 56.5 J/K mol

S°f for NaCl(s) = 72.13 J/K mol

[

(

)

(

)]

+ ∆S°= 1 mol Na 0.0590 kJ + 1 mol Cl 0.0565 kJ mol rxn K mol Na+ mol rxn K mol Cl-

–

3.

Instructions indicate that you should calculate ΔG° using the answers to (a) and (b). Use the Gibbs- Helmholtz equation.

It is important to ensure that the units are consistent throughout the equation. In other words, you must convert ΔS ° to kJ/K mol before substituting the values into the Gibbs equation.

In general, if a question requiring the Gibbs Equation does not provide a specific temperature, assume SATP conditions (or T = 25°C = 298 K) apply. The presence of the ° symbol, however, does not restrict all calculations to a temperature of 25°C.

4.

A (–) ΔG value indicates spontaneity (also called thermodynamic favorability) though the magnitude is not very large.

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[

(

1 mol NaCl 0.07213 kJ mol rxn K mol NaCl

)]

= 0.04337 kJ/K mol rxn

(c) ∆G° = ∆H° − T∆S° ∆G° = 3.87 kJ – 298 K 0.04337 kJ molrxn K molrxn = –9.05 kJ/molrxn

(

)

(d) The dissolving of the salt has a (–) ∆G value meaning it is spontaneous as would be expected for dissolving a salt at room temperature.

In fact, endothermicity and an increased number of microstates also make sense for the dissolving process. This process is entropy driven.

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It is significant that ΔS° values are usually very small compared to ΔH°. Only multiplication by T causes subtraction of the TΔS° term from ΔH° to have a significant impact on the value of ΔG°. In an example such as the dissolving of table salt, both ΔH° and ΔS° are positive. This dissolution process is spontaneous only when the temperature is large enough to allow subtraction of the TΔS° term from ΔH° to result in a negative ΔG°.

The values of standard free energies of formation of pure compounds and elements and aqueous ions are listed in Table A8 in the appendix. As is the case for the S° and ΔHf° values in the appendix, standard conditions for ΔGf° are 25°C and 1 atm.

Figure 3.2.4 Examination of the equation, ΔG° = ΔH° – TΔS°,

shows that if ΔH° and ΔS° share the same sign, temperature will determine spontaneity. The temperature at which ΔH° and the term TΔS° are equal is the temperature at which equilibrium will occur. “High” and “low” temperatures are relative and refer to temperatures above or below the equilibrium T value.

Sample Problem 3.2.1(b) — Using Free Energies of Formation to Calculate a Free Energy Change at Standard Temperature For the process NaCl(s) → Na+(aq) + Cl–(aq), use the tables in Table A8 in the appendix to calculate: (a) ΔG° (b) According to your results in (a), is this a spontaneous process?

What to Think About 1.

As ΔG° is a state function, calculate the value using the equation introduced earlier in this section. The ΔG°f values are listed in Table A8 in the appendix.

As with ΔH°f values, elements in their standard states have ΔG°f values of 0 kJ/mol. This is not the case for ions in solution.

The value calculated for ΔG° using reference values and the state function equation is very close to the value calculated using the Gibbs-Helmholtz equation. Using the state function algorithm to calculate ΔH° and ΔS° values introduces further approximations due to rounding to appropriate significant figures. This leads to differences between the two calculated values.

2.

A (–) ΔG value indicates spontaneity even though the magnitude is not very large.

196 Chapter 3 Thermodynamics

How to Do It

(a) ∆G° = Σn∆G°(prod) – Σn∆G°(react)

∆G°f for Na+(aq) = –261.91 kJ/mol ∆G° for Cl-(aq) = –131.23 kJ/mol f

∆G°f for NaCl(s) = –384.12 kJ/mol

[ [

(

)

(

)]

+ kJ + 1 mol Cl- -131.23 kJ ∆G° = 1 mol Na -261.91 + mol rxn mol Clmol rxn mol Na

(

)]

1 mol NaCl –384.12 kJ mol rxn mol NaCl

= –9.02 kJ/mol rxn

b) This is a spontaneous process. This makes sense for the dissolving of a salt at room temperature. We know that virtually all salts dissolve spontaneously at room temperature. This is generally entropy driven. It is interesting that the dissolving of NaCl is very nearly at equilibrium under SATP conditions.

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Practice Problems 3.2.1 — Using the Gibbs-Helmholtz Equation and Free Energies of Formation to Calculate a Free Energy Change at Standard Temperature 1.

Use the ΔGf° values in Table A8 in the appendix to calculate ΔG°comb for the combustion of ethanol under SATP conditions.

2.

Calculate the maximum work available from the oxidation of 1 mole of octane at 25°C and 1 atm. Use the ΔH°f and the S°f values from Table A8 in the appendix along with the Gibbs-Helmholtz equation.

3.

Calculate ΔG°rxn for the reaction of N2(g) + O2(g) → 2 NO(g). Do the calculation two ways. First by determining ΔH°f , then ΔS°f followed by the Gibbs-Helmholtz equation. Then use the ΔG°f values and see how the values compare.

Calculation of Free Energy at Non-Standard Temperatures

ΔG° is the most temperature dependent of the thermodynamic functions. We can use the GibbsHelmholtz equation to calculate ΔG° at temperatures other than 25°C. To do this, we neglect the variations of ΔH° and ΔS° with temperature because they are relatively temperature independent quantities compared to ΔG°. ΔH° is essentially independent of temperature and the variations of ΔS° with temperature are ordinarily very small — entropy is most dependent on temperature near absolute zero. Near room temperature, variations in ΔS° are negligible.

In other words, we take the values of ΔH° and ΔS° from the tables and simply insert them into the Gibbs-Helmholtz equation using the appropriate value of T (in Kelvin degrees) to calculate ΔG°.

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Sample Problem 3.2.2 — Using the the Gibbs-Helmholtz Equation to Calculate a Free Energy Change at Non-Standard Temperatures Calculate ΔG° for the reaction Cu(s) + H2O(g) → CuO(s) + H2(g) at a temperature of 227.0°C.

What to Think About 1.

2.

3.

As the temperature is non-standard, you must use the Gibbs-Helmholtz equation to solve the problem.

How to Do It

∆H°(overall) = Σ n∆H°f(prod) – Σ n∆H°f(react) ∆H°f for CuO (s) = –157.3 kJ/mol ∆H°f for H2 (g) = 0 kJ/mol

Use reference tables to determine the ΔH° and ΔS° values of each compound or element in the reaction first.

∆H°f for Cu (s) = 0 kJ/mol

Calculate the overall enthalpy change for the reaction.

0 kJ ∆H° = 1 mol CuO –157.3 kJ + 1 mol H2 mol rxn mol CuO mol H2 mol rxn

∆H°f for H2O (g) = –241.82 kJ/mol

[

[

– 1 mol Cu mol rxn

(

(

)

Evaluate the sign of the enthalpy value and its significance.

5.

Calculate the overall entropy change for the reaction

(

(

0 kJ + 1 mol H2O mol Cu mol rxn

= 84.5 kJ/molrxn 4.

)

)] )]

–241.82 kJ 1 mol rxn

A slightly endothermic process ∆S° = ΣnS°(prod) – ΣnS°(react) S°f for CuO(s) = 42.63 J/K mol S°f for H2(g) = 130.68 J/K mol S°f for Cu(s) = 33.15 J/K mol S°f for H2O(g) = 188.83 J/K mol

[

(

)

(

∆S° = 1 mol CuO 42.63 kJ + 1 mol H2 130.68 kJ mol rxn K mol CuO K mol H2 mol rxn

[

(

)

(

– 1 mol Cu 33.15 kJ + 1 mol H2O 188.83 kJ mol rxn K mol Cu mol rxn K mol rxn = –48.67 kJ/molrxn K 6.

Consider the sign of the overall entropy change of the reaction and its significance.

7.

Finally, use the Gibbs-Helmholtz equation to calculate ΔG° at a temperature of 227.0 °C, but first convert the temperature to Kelvin degrees.

)]

)]

A (–) value for ∆S makes sense as the O atoms have moved from a gaseous compound to a solid compound. Additionally the H has moved from gaseous water where it was combined with oxygen to pure gaseous hydrogen.

( )

227 °C 1K + 273 K = 500.K 1°C ∆G° = ∆H° − T∆S° ∆G° = 84.5 kJ – 500. K –0.04867 kJ K molrxn molrxn

(

)

= 108.8 kJ/molrxn Note that the (+) ∆G° value indicates a non-spontaneous reaction. The (+) ∆H° and the (–) ∆S° both favor a nonspontaneous reaction. Hence this reaction would be nonspontaneous under any temperature conditions.

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Practice Problems 3.2.2 — Using the Gibbs-Helmholtz Equation to Calculate a Free Energy Change at Non-Standard Temperatures 1.

The mineral calcite decomposes according to the following reaction: CaCO3(s) → CaO(s) + CO2(g) in which ΔH° = 178.3 kJ/molrxn and ΔS° = 160.6 J/molrxn K. (a) Calculate ΔG° at 750.0°C.

(b) Calculate ΔG° at 950.0°C.

(c) Does the reaction become more or less spontaneous (thermodynamically favorable) as temperature increases?

Experimental Manipulation of the Gibbs-Helmholtz Free Energy Equation

We can arrange the equation, G = H – TS to the form G = –ST + H. A plot of G vs. T will be a straight line with a slope of −S and a y-intercept of H. Various experiments may be set up to study the variables in the Gibbs-Helmholtz equation. These experiments produce graphs such as those shown in Figure 3.2.5.

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� Figure 3.2.5 The graph above shows that free energy (G) varies with temperature (T).

Since all chemical species have a positive value for entropy, G decreases with an increase in temperature (at constant pressure and state). Gases have far greater entropy than liquids, which have greater entropy than solids. Figure 3.2.5 shows the free energies of gases are more temperature dependent than liquids, which are more temperature sensitive than solids.

The direct measurement of free energy is not easily accomplished. In the next section, we will apply our understanding of equilibrium to a more detailed investigation of free energy and how it varies with temperature.

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3.2 Review Questions 1.

(a) What two factors, other than temperature, influence the spontaneity of a process?

(b) Describe each factor.

(c) What sign is associated with spontaneity for each factor?

2.

3.

Complete the following table for the equation: ΔG° = ΔH° − TΔS°.

Sign of ∆H°

Sign of ∆S°

Temperature

–

+

low

–

+

high

+

–

low

+

–

high

–

–

+

+

–

–

+

+

Sign of ∆G°

Spontaneous /Non-Spontaneous

– spontaneous high +

Examine the representation of ice melting and water condensing on a glass of ice water. Consider circled areas A and B. Circle A: (i) What are the signs of ΔH° and ΔS° for this process? Explain your answers.

(ii) Is this process spontaneous at all temperatures, only low temperatures, only high temperatures or never? Explain.

Circle B: (i) What are the signs of ΔH° and ΔS° for this process? Explain your answers.

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(ii) Is this process spontaneous at all temperatures, only low temperatures, only high temperatures or never? Explain.

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ΔH° is –196.1 kJ/molrxn for the following representation of a reaction of H2O2(l).

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(a) What are the signs of ΔS° and ΔG° for the reaction? Explain.

(b) Is the process spontaneous at all temperatures, only low temperatures, only high temperatures or never? Explain.

5.

(a) Use the representation below to produce a balanced chemical equation. Use the lowest possible whole number coefficients in your equation.

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(b) Use Table A8 in the appendix to calculate ΔH°rxn for the reaction. Show your work.

(c) Predict the sign of ΔS°rxn for the reaction (do not calculate).

(d) Is this process spontaneous at all temperatures, only low temperatures, only high temperatures or never? Explain.

(e) Determine the temperature (in °C) at which the spontaneity of this reaction changes.

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Shiny silver iodine crystals are deposited on the surface of ice placed into a test tube filled with purple iodine vapor. (a) Write a balanced chemical equation to describe the process pictured here.

(b) Use Table A8 in the appendix to calculate the sublimation temperature (same as the deposition temperature) for iodine.

7.

Consider the following reaction: 2 O3(g) → 3 O2(g). ΔH° = –285.4 kJ/molrxn. (a) Predict the signs for ΔS°rxn and ΔG°rxn and the spontaneity of the reaction (no calculations required).

(b) Use values from Table A8 in the appendix to calculate ΔS°rxn and ΔG°rxn and to determine the accuracy of your prediction.

8.

Calculate ΔG°rxn for each of the following reactions using Table A8 in the appendix. (a) NH3(g) + HCl(g) → NH4Cl(s)

(b) AgNO3(aq) + NaI(aq) → AgI(s) + NaNO3(aq) (Hint: Use the net ionic equation.)

(c) P4O10(s) + H2O(l) → H3PO4(l)

9.

(Hint: The reaction is unbalanced.)

Government requirements in the United States and Canada state that fuels should contain a “renewable” component.

Fermentation of glucose isolated from a variety of garden products such as sugar cane, beets, and corn produces ethanol.

Producers can add ethanol to gasoline to fulfill the renewable fuel requirement. Here is the fermentation reaction for glucose: C6H12O6(s) → 2 C2H5OH(l) + 2 CO2(g)

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Calculate: (a) ΔH°

(b) ΔS°

(c) ΔG°

(d) Under what temperature conditions does this reaction occur spontaneously? Justify.

(e) How might this impact the economic feasibility of producing renewable fuels in this way?

10. In low concentrations, phosgene is a colorless to pale yellow cloud with the pleasant odor of freshly cut hay. In higher concentrations, phosgene becomes a choking agent toxic to the pulmonary system. Armies used phosgene to kill hundreds of people during the World War 1. Carbon monoxide and chlorine combine to form phosgene, COCl2(g). (a) Calculate ΔS°rxn given that ΔH° = –220. kJ/molrxn and ΔG° = –206 kJ/molrxn at 25 °C.

(b) The (–)ΔG° value means the production of the war gas is spontaneous under SATP conditions. Considering the signs of ΔS° and ΔH°, what temperature conditions should further favor the spontaneous production of phosgene?

(c) Assuming the temperature effects on ΔS° and ΔH° are negligible, calculate ΔG° at 435°C. Is the reaction spontaneous at this higher temperature?

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3.3 Pressure, Equilibrium, and Free Energy Warm Up In this activity, you will determine if entropy is increased or decreased when you stretch a balloon. Your determination will help you complete the ΔS cells in the table below.

• Stretch a balloon (or a thick rubber band) by pulling and relaxing multiple times.

• Hold the balloon against your upper lip or forehead during a stretch. Keep it in contact with your face while letting the balloon relax slowly.

1.

What do you feel when the balloon is stretched? When it is relaxed? _____________________________________________________________

2.

Complete the ΔH portion of the data table below using + or – signs. ΔH

ΔG

ΔS

Representation

Stretched Relaxed 3.

Does the balloon stretch or relax spontaneously? Stretch the balloon and let go of the both ends simultaneously to find out. _____________________________________________________________

4.

Complete the ΔG portion of the table.

5.

What MUST the sign of ΔS be? Indicate in the table above.

6.

Is the relaxation of the balloon enthalpy or entropy driven? _____________________________________________________________

7.

Draw a molecular representation to show how the chains of molecules must align themselves when you stretch the balloon and when you relax it. (Rubber is a polymer, which consists of long chains of repeating chemical units called monomers.

See section 9.3 of AP Chemistry 1 for a discussion of polymers.

8.

Do the sign of ΔS and the drawings of chain alignment representation portions of your table agree? Why or why not? ________________________________________________________________________________________________ ________________________________________________________________________________________________

204 Chapter 3 Thermodynamics

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Dependence of Free Energy on Pressure

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The entropy of 1 mole of a gas in a 20.0 L container is larger than the entropy of 1 mole of the same gas in a 10.0 L container. There are more microstates for the gas or ways for the molecules of the gas to arrange themselves in the space in a larger volume. If the entropy of a gas depends on the volume and consequently the pressure, then the free energy of the gas must also depend on these two factors.

Experimentally, it can be seen that: G = G° + RT ln P where G° is the free energy of a gas sample at 1 atm pressure, and G is the free energy of the gas at some pressure P. R is 8.314 J/mol.K, and T is the temperature in Kelvin degrees.

How does the free energy change for a reaction depend upon pressure?

Methane “burns” in oxygen to form carbon dioxide and water: CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g) For this reaction: ΔG = GCO + 2 GH O – GCH4 – 2 GO (under non-standard conditions) 2

2

2

If we use G = G° + RT ln P for each component: ΔG = (G°CO + RT ln PCO ) + 2(G°H O + RT ln PH O) – (G°CH + RT ln PCH ) 2

2

2

2

4

4

– 2(G°O + RT ln PO ) 2

2

Collecting all the G° terms and all the RT ln P terms: ΔG = G°CO + 2 G°H O – G°CH – 2 G°O + RT (ln PCO + 2 ln PH O – ln PCH – 2 ln PO ) 2

2

4

2

2

2

4

2

Since ΔG° = G°CO2 + 2 G°H O – G°CH – 2 G°O and 2 ln x = ln x2 2

4

2

We have ΔG = ΔG°+ RT (ln PCO + ln P 2H O – ln PCH – ln P 2O ) 2

2

4

2

Applying log rules: ln a + ln b = ln (a ⋅ b) and ln c – ln d = ln (c/d), we end up with ΔG = G° + RT In

(P )(P ) (P )(P ) CO2

CH4

2

H2O 2

O2

The expression in square brackets is the reaction quotient Q from section 2.5.

You do not need to be able to derive this equation. We provide the mathematical justification for those who want to understand how the reaction quotient appears in the equation when we consider a change in the Gibbs free energy of a reaction. For any reaction with species present at nonstandard pressures: ΔG = ΔG° + RT ln Q You can calculate ΔG° for a reaction from reference values of free energies of formation of reactants and products as in section 3.2. Using ΔG° and Q, you can determine ΔG. The sign of ΔG allows you to predict whether the reaction will proceed spontaneously forward (– sign) or reverse (+ sign) to reach equilibrium. If a reaction mixture is at equilibrium ΔG will be 0.

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Sample Problem 3.3.1 — Calculation and Use of ΔG Consider the reaction: 2 NO2(g)  N2O4(g)

ΔG° = –5.39 kJ/molrxn at 25°C

A mixture contains NO2 with a partial pressure of 0.10 atm and N2O4 with a partial pressure of 0.20 atm. Use ΔG to determine whether the gas mixture is at equilibrium. If it is not, in which direction will the reaction proceed?

What to Think About

How to Do It

1.

The pressures are not 1 atm and the question asks for the value of ΔG as opposed to ΔG°, so the equation required is: ΔG = ΔG ° + RT ln Q

Determine the reaction quotient Q by placing the partial pressure of the product over the partial pressure of the reactant and raising the pressure of each species to the power of its coefficient: PN O Q = 2 2 4 = 0.20 atm = 20 (0.10 atm)2 P NO2

2.

The temperature must be expressed in Kelvin degrees.

25°C x 1 K + 273 K = 298 K 1°C

3.

The value of the gas constant must have units of energy. Recall: 0.08206 L·atm × 101.3 J = 8.31 J mol·K L·atm mol·K

If K were given in the question, a comparison of Q to K would predict the reaction’s preferred direction.

However, K is not given, hence

(or more precisely 8.314) 4.

Be sure the G° and R units are the same (both kJ or both J).

∆G = ∆G° + RT ln Q

(

)

= –5.39 kJ + 0.008314 kJ x 298 K x ln(20.) molrxn mol.K = –5.39 kJ/molrxn + 7.42 kJ/mol = 2.03 kJ/molrxn or 2.0 kJ/molrxn

A non-zero ∆G value indicates the system is not at equilibrium.

5.

Be sure to carry the correct signs through the entire calculation.

206 Chapter 3 Thermodynamics

The system is NOT at equilibrium. Because the sign of ∆G is positive, the reaction will proceed in the reverse direction (to the LEFT) to reach equilibrium.

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Practice Problems 3.3.1 — Calculation and Use of ΔG 1.

What is the value for Gibbs free energy in the production of ammonia from nitrogen and hydrogen: N2(g) + 3 H2(g) 2 NH3(g), if the pressure of all three gases is 2.00 atm and the temperature is 23°C? Begin by determining ΔG° for this reaction using the reference values from Table A8 in the appendix.

2.

ΔG° for the reaction H2(g) + I2(g)  2 HI(g) is 2.60 kJ/molrxn at 25°C. Calculate ΔG and predict the direction in which the reaction proceeds spontaneously if the initial pressures are 7.0 atm, 3.0 atm, and 3.50 atm for H2, I2, and HI respectively.

3.

Methanol, a common gasoline additive is also present as a component in windshield washer solution. A common method for synthesis of this alcohol is to react carbon monoxide with hydrogen gas: CO(g) + 2 H2(g)  CH3OH(l) Calculate ΔG for this reaction at 25°C when 2.5 atm of carbon monoxide and 1.5 atm of hydrogen gases react to form liquid methanol. Begin by determining ΔG° for this reaction using the reference values from Table A8 in the appendix.

The relationship between the reaction quotient and free energy is:

Free Energy and the Equilibrium Constant

ΔG = ΔG° + RT ln Q At equilibrium, ΔG = 0. The reaction mixture has reached a minimum free energy. Also at equilibrium, Q = Keq , so the reaction quotient is equal to the equilibrium constant.

Then, at equilibrium, ΔG = ΔG° + RT ln Q becomes Thus,

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0 = ΔG° + RT ln K ΔG° = –RT ln K

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With this expression, we can use the reference values of ΔG°f to calculate ΔG° for a reaction and then calculate K. Note 1: Some references prefer to deal with logs to the base 10, rather than natural logs to the base e.

In that case, since 2.303 log x = ln x, the above expression becomes: ΔG ° = –2.303 RT log K Note 2: Usually the change in Gibbs free energy at equilibrium is calculated from the partial pressures of the reagents in a reaction such as the oxidation of methane gas, so the equilibrium constant is actually Kp. However, occasionally the change in Gibbs free energy at equilibrium may refer to Kc in situations such as low solubility salt equilibria (Ksp ) or weak acids (Ka ).

Using ΔG° and K to Predict Reaction Direction

There are three possible situations when calculating the change in Gibbs free energy at equilibrium: 1. If ΔG° is negative, ln K must be positive. This means that K is greater than 1 and the reaction proceeds spontaneously (is thermodynamically favorable) in the forward direction.

2. If ΔG° is positive, ln K must be negative. This means that K is less than 1 and the reverse reaction is spontaneous.

3. If ΔG° = 0, ln K = 1. The reaction is at equilibrium.

If ΔG° is a large negative number, K will be much greater than one. This means that the forward reaction will go virtually to completion. Conversely, if ΔG° is a large positive number, K will be a tiny fraction, much less than 1, and the reverse reaction will go nearly to completion. Table 3.3.1 summarizes these relationships.

Table 3.3.1 ∆G°, K, and Reaction Direction

ΔG° negative

K>1

spontaneous reaction

ΔG° = zero

K=1

at equilibrium

ΔG° positive

K<1

reverse reaction is spontaneous

ΔG° <<< 0

K very BIG

spontaneous to near completion

ΔG° >>> 0

K nearly 0

reverse reaction to near completion

Sample Problem 3.3.2(a) — Calculating ΔG° from K Values The equilibrium constant K for the reaction PCl3(g) + Cl2(g)  PCl5(g) is 24.0 at 250.°C. Calculate ΔG°. Is the reaction spontaneous in the forward or reverse direction?

What to Think About

How to Do It

1.

Convert the temperature to Kelvin degrees.

250.°C x 1 K + 273 K = 523 K 1°C

2.

Use the equation ΔG° = –RT ln K to calculate ΔG°.

3.

It is critical to use the appropriate R value and to carry the negative sign that precedes the equation through each step in the equation.

∆G° = –RT ln K = – 8.314 J x 523 K x ln (24.0) mol·K

4.

If K > 1, ΔG° must be negative.

208 Chapter 3 Thermodynamics

= –13 800 J/molrxn = –13.8 kJ/molrxn The forward reaction is spontaneous.

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Sample Problem 3.3.2(b) — Conversion of ΔG° to K Phosphorus pentachloride decomposes as follows: PCl5(g) PCl3(g) + Cl2(g). ΔG° = 26.5 kJ/molrxn at 227°C. Calculate the equilibrium constant, K. Does the reaction favor products or reactants under these conditions?

What to Think About

How to Do It

1.

Convert the temperature to Kelvin degrees.

227°C x 1 K + 273 K = 500. K 1°C

2.

As ΔG° and K are in the question, use the equation ΔG° = –RT ln K.

∆G° = –RT ln K

3.

4.

It is critical to use the appropriate R value and to carry the negative sign that precedes the equation throughout the calculations.

Make sure the energy units for ΔG° and R agree.

so ln K = –∆G° RT hence, K = e

∆G° − RT

26.5 kJ/molrxn where ∆G° = = –6.37 –RT –0.008314 kJ x 500. K mol.K Thus K = e-6.37 = 1.70 x 10–3

Practice Problems 3.3.2 — Conversions of ΔG° to K and K to ΔG° 1.

The pollutant gas nitrogen monoxide forms from nitrogen and oxygen gas according to the following equation: N2(g) + O2(g) 2 NO(g). K is only 8.44 x 10–3 at 2500°C. Calculate ΔG°. Is the reaction spontaneous in the forward or the reverse direction?

2.

(a) Calculate K for a reaction at 25°C that has a ΔG° value of –35.0 kJ/molrxn.

(b) Repeat the calculation at the same temperature for a reaction with ΔG° equal to +35.0 kJ/molrxn.

(c) Compare the K values and what they mean for these two reactions.

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In an exergonic reaction (one in which ΔG° is negative) the products (point 2 in Figure 3.3.1(a)) have less free energy and are more stable than the reactants (point 1 in Figure 3.3.1(a)). The difference between these two points (2 – 1) is ΔG°. All points between 1 and 2 correspond to various mixtures of the reactants and products present as the reaction proceeds. The difference between point 1 (in the forward direction) or point 2 (in the reverse direction) and any point on the curve is ΔG. The mixture of reactants and products present at the minimum of the curve (point 3) has the least free energy and is the most stable of all. This represents the equilibrium mixture.

At any point on the curve, a comparison of the reaction quotient Q to the equilibrium constant K indicates the direction in which the reaction must proceed to reach equilibrium. This will always be the direction that is downhill in a free energy diagram! For an exergonic reaction, because ΔG° is negative, K must be > 1 so the equilibrium mixture will always contain more products than reactants.

In an endergonic reaction (one in which ΔG° is positive) the products (point 2 in Figure 3.3.1(b)) will have more free energy than the reactants (point 1 in Figure 3.3.1(b)). In this case, K < 1 and the standard reaction is reactant favored, resulting in an equilibrium mixture containing more reactants than products.

Variations in Total Free Energy for Reversible Reactions

G vs. Reaction Progress for an Exergonic Reaction

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210 Chapter 3 Thermodynamics

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3.3 Review Questions 1.

Calculate ΔG° for the conversion of oxygen to ozone: 3 O2(g) 2 O3(g) at 25°C given that K = 2.50 x 10–29

2.

Consider the reaction: MgCO3(s) MgO(s) + CO2(g) (a) Is the forward reaction spontaneous at room temperature (25°C)? Give a mathematical justification.

(b) Is the reaction driven by enthalpy, entropy, both, or neither? Give a mathematical justification.

(c) What is the value of K at 25°C?

(d) At what temperature does the direction of the spontaneous reaction change?

(e) Does high temperature make the reaction more or less spontaneous in the forward direction? Include a full explanation.

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Methane, CH4(g), is the main component of natural gas. When the barrel of a Bunsen burner is wide open, more oxygen (O2(g)) is available for the combustion of the methane. The products are entirely CO2(g) and H2O(g). When the barrel is nearly closed, the products are CO2(g) and some H2O(l).

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(a) Calculate ΔG° at 25°C for each reaction.

(b) Calculate K for each reaction.

(c) Which burner adjustment gives a hotter flame?

(d) Why does opening the barrel on the burner force a more complete combustion?

4.

Use the reference values for ΔH°f and S° from Table A8 in the appendix to calculate ΔH°, ΔS°, and ΔG° for the reaction: CO(g) + H2O(g) CO2(g) + H2(g) at 25°C. (a) ΔH° (Show all work completely.)

(b) ΔS° (Show all work completely.)

(c) ΔG° (Use the Gibbs-Helmholtz equation with a temperature of 25°C.)

212 Chapter 3 Thermodynamics

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(d) Calculate the equilibrium constant K for this reaction at 25°C.

(e) Calculate the equilibrium constant K for this reaction at 225°C.

(f) Using your answers to parts (d) and (e), determine whether this reaction is exo- or endothermic? Justify your answer. Does this response agree with the answer to part (a)?

5.

Consider what each of the following graphs represent.

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(a) How are the graphs similar? Be specific.

(b) How are the graphs different? Be specific.

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Consider the following representations of a physical and a chemical change. Answer each question for A and B.

� A. A physical change

B. A chemical change

(a) What is the sign of ΔH°?

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(b) What is the sign of ΔS°?

(c) How does the sign of ΔG° vary with temperature?

7.

Consider the following reaction:

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(a) Calculate ΔG°.

(b) Examine the following molecular representation. Assume each molecule exerts a partial pressure of 0.010 atm at 25°C. Calculate ΔG.

(c) In general, how would the representation change if the system were to achieve equilibrium?

214 Chapter 3 Thermodynamics

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Consider the following reaction at 25°C. 2 NO(g) + O2(g) 2 NO2(g) ΔG° = –71.2 kJ/molrxn (a) Calculate ΔG under the following conditions: PNO = 0.0500 atm PO = 0.0500 atm PNO = 0.100 atm 2

2

(b) Is the reaction more or less spontaneous under these pressure conditions?

9.

(a) Calculate ΔG at 298 K for the reaction 2 ICl(g) I2(g) + Cl2(g) in a mixture containing the following partial pressures: PICl = 2.0 atm PI = 0.0020 atm PCl = 0.0020 atm 2

2

(Remember: I2 is normally a solid under standard conditions.)

(b) Is this reaction more or less spontaneous under these conditions than under standard conditions?

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10. At 25°C, acetic acid, CH3COOH, ionizes in water to form hydrogen and acetate ions. The acid is quite weak and has a small equilibrium constant, Ka, of 1.8 x 10–5. CH3COOH(aq) H+(aq) + CH3COO–(aq) (a) Calculate ΔG° for the ionization of acetic acid.

(b) Calculate ΔG for the reaction if [CH3COOH] = 0.10 mol/L, [H+] = 9.5 x 10–5 mol/L, and [CH3COO–] = 9.5 x 10–5 mol/L.

(c) Compare the spontaneity of ionization under the conditions in part (b) to those under standard conditions.

216 Chapter 3 Thermodynamics

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